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let $f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x $ \epsilon R $ and

let $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-ln t\right) f(t) dt,$ for all x $ \epsilon (1,\infty)$

which of the following is true?

A) g is increasing on $(1, \infty)$

B) g is decreasing on $(1, \infty)$

C) g is increasing on (1, 2) and decreasing on $(2,\infty)$

D) g is decreasing on (1, 2) and increasing on $(2,\infty)$

Option B

Concept involved use of Newton leibnits formula

i.e, $\frac{d}{dx}\left\{\int_{g(x)}^{f(x)} \phi(t) dt\right\}=\phi (f(x)).f'(x)-\phi (g(x)).g'(x)$

Sol. Here , $f(x)(1-x)^{2} \sin^{2}x+x^{2},$ for all x

$g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log(t)\right)f(t)dt$

$\forall x \epsilon (1, \infty)$

Here,

$f(x)=(1-x)^{2}\sin^{2} x+x^{2}$ for all x

and $g(x)=\int_{1}^{x} \left(\frac{2(t-1)}{t+1}-\log t \right) f(t) dt,$

$\Rightarrow$ $g'(x)=\left\{\frac{2(x-1)}{x+1}-\log x\right\}f(x) (+ve)....(i)$

for g'(x) to be increasing or decreasing

Let $\phi(x)=\frac{2(x-1)}{(x+1)}-\log x$

$ \phi'(x)= \frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{x(x+1)^{2}}$

$\phi '(x) <0,$ for x >1

$\Rightarrow$ $\phi (x) < \phi(1) \Rightarrow \phi (x) <0$......(ii)

For Eqs.(i) and (ii) , we get

$g'(x) <0$ for $ x \epsilon (1 ,\infty)$

$\therefore$ g(x) is decreasing for $ x \epsilon (1 ,\infty)$

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